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Table of contents
About eBook PLUS vChapter 1 — Graphs and polynomials 1 Exercise 1A — The binomial theorem 1 Exercise 1B — Polynomials 2
Exercise 1C — Division of polynomials 4 Exercise 1D — Linear graphs 7
Exercise 1E — Quadratic graphs 8 Exercise 1F — Cubic graphs 11 Exercise 1G — Quartic graphs 13 Chapter review 15
Short answer 15 Multiple choice 16 Extended response 17
Chapter 2 — Functions and transformations 21 Exercise 2A — Transformations and the parabola 21 Exercise 2B — The cubic function in power form 22 Exercise 2C — The power function (the hyperbola) 25 Exercise 2D — The power function (the truncus) 29 Exercise 2E — The square root function in power form 32 Exercise 2F — The absolute value function 34
Exercise 2G — Transformations with matrices 38 Exercise 2H — Sum, difference and product functions 40 Exercise 2I — Composite functions and functional
equations 41 Exercise 2J — Modelling 43 Chapter review 46 Short answer 46 Multiple choice 48 Extended response 49 Exam practice 1 52 Short answer 52 Multiple choice 52 Extended response 53
Chapter 3 — Exponential and logarithmic equations 54 Exercise 3A — The index laws 54
Exercise 3B — Logarithm laws 55 Exercise 3C — Exponential equations 56 Exercise 3D — Logarithmic equations using
any base 58
Exercise 3E — Exponential equations (base e) 60 Exercise 3F — Equations with natural (base e)
logarithms 62 Exercise 3G — Inverses 63 Exercise 3H — Literal equations 65
Exercise 3I — Exponential and logarithmic modelling 66 Chapter review 68
Short answer 68 Multiple choice 69 Extended response 70
Chapter 4 — Exponential and logarithmic graphs 72 Exercise 4A — Graphs of exponential functions with
any base 72
Exercise 4B — Logarithmic graphs to any base 76 Exercise 4C — Graphs of exponential functions with
base e 81
Exercise 4D — Logarithmic graphs to base e 86 Exercise 4E — Finding equations for graphs of exponential
and logarithmic functions 90
Exercise 4F — Addition of ordinates 91
Exercise 4G — Exponential and logarithmic functions with absolute values 95
Exercise 4H — Exponential and logarithmic modelling using graphs 96
Chapter review 98 Short answer 98 Multiple choice 99 Extended response 100 Chapter 5 — Inverse functions 103
Exercise 5A — Relations and their inverses 103 Exercise 5B — Functions and their inverses 105 Exercise 5C — Inverse functions 109
Exercise 5D — Restricting functions 111 Chapter review 117
Short answer 117 Multiple choice 120 Extended response 121
Chapter 6 — Circular (trigonometric) functions 123 Exercise 6A — Revision of radians and the unit circle 123 Exercise 6B — Symmetry and exact values 124
Exercise 6C — Trigonometric equations 127 Exercise 6D — Trigonometric graphs 130 Exercise 6E — Graphs of the tangent function 135 Exercise 6F — Finding equations of trigonometric
graphs 137
Exercise 6G — Trigonometric modelling 138 Exercise 6H — Further graphs 139
Exercise 6I — Trigonometric functions with an increasing trend 144 Chapter review 145 Short answer 145 Multiple choice 147 Extended response 148 Exam practice 2 148 Short answer 148 Multiple choice 150 Extended response 151 Chapter 7 — Differentiation 152
Exercise 7A — Review — gradient and rates of change 152 Exercise 7B — Limits and differentiation from first
principles 154
Exercise 7C — The derivative of xn 158 Exercise 7D — The chain rule 160 Exercise 7E — The derivative of ex 164 Exercise 7F — The derivative of loge(x) 167 Exercise 7G — The derivatives of sin(x), cos(x) and
tan(x) 172
Exercise 7H — The product rule 176 Exercise 7I — The quotient rule 179
Exercise 7J — Mixed problems on differentiation 183 Chapter review 190
Short answer 190 Multiple choice 192 Extended response 195
Chapter 8 — Applications of differentiation 196 Exercise 8A — Equations of tangents and normals 196 Exercise 8B — Sketching curves 199
Exercise 8C — Maximum and minimum problems when the function is known 209
Exercise 8D — Maximum and minimum problems when the function is unknown 210
Exercise 8E — Rates of change 213 Exercise 8F — Related rates 215 Exercise 8G — Linear approximation 217 Chapter review 217 Short answer 217 Multiple choice 219 Extended response 221 Chapter 9 — Integration 225 Exercise 9A — Antidifferentiation 225
Exercise 9B — Integration of ex, sin(x) and cos(x) 230 Exercise 9C — Integration by recognition 233 Exercise 9D — Approximating areas enclosed by
functions 237
Exercise 9E — The fundamental theorem of integral calculus 238
Exercise 9F — Signed areas 244 Exercise 9G — Further areas 249
Exercise 9H — Areas between two curves 260 Exercise 9I — Average value of a function 270 Exercise 9J — Further applications of integration 271 Chapter review 276 Short answer 276 Multiple choice 278 Extended response 281 Exam practice 3 283 Short answer 283 Multiple choice 284 Extended response 285
Chapter 10 — Discrete random variables 286 Exercise 10A — Probability revision 286 Exercise 10B — Discrete random variables 289 Exercise 10C — Measures of centre of discrete random
distributions 293
Exercise 10D — Measures of variability of discrete random distributions 297
Chapter review 302 Short answer 302 Multiple choice 304 Extended response 306
Chapter 11 — The binomial distribution 308 Exercise 11A — The binomial distribution 308 Exercise 11B — Problems involving the binomial
distribution for multiple probabilities 311 Exercise 11C — Markov chains and transition
matrices 315
Exercise 11D — Expected value, variance and standard deviation of the binomial distribution 318 Chapter review 321
Short answer 321 Multiple choice 323 Extended response 325
Chapter 12 — Continuous distributions 328 Exercise 12A — Continuous random variables 328 Exercise 12B — Using a probability density function to find
probabilities of continuous random variables 330 Exercise 12C — Measures of central tendency and
spread 336
Exercise 12D — Applications to problem solving 341 Exercise 12E — The normal distribution 343 Exercise 12F — The standard normal distribution 347 Exercise 12G — The inverse cumulative normal
distribution 350 Chapter review 354 Short answer 354 Multiple choice 357 Extended response 358 Exam practice 4 360 Short answer 360 Multiple choice 361 Extended response 361 Solutions to investigations 363 Chapter 1 363
Investigation — Quartics and beyond 363 Chapter 2 363
Investigation — Goal accuracy 363 Chapter 9 363
Investigation — Definite integrals 363 Chapter 11 — 363
Investigation — Winning at racquetball! 363 Chapter 12 364
G r a p h s a n d p o l y n o m i a l s M M 1 2 - 1
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Exercise 1A — The binomial theorem
1 a (x + 3)2= 2 2 0 x + 1 1 2 3 1 x + 2 2 3 2 = x2+ 6x + 9 b (x + 4)5 = 5 5 0 x + 4 5 4 1 x + 3 2 5 4 2 x + 2 3 5 4 3 x + 4 5 4 4 x + 5 5 4 5 = x5+ 5x44 + 10x316 + 10x264 + 5x256 + 1 × 1024 = x5+ 20x4+ 160x3+ 640x2+ 1280x + 1024 c (x − 1)8 = 8 8 0 x + 7 8 ( 1) 1 x − + 6 2 8 ( 1) 2 x − + 8 5( 1)3 3 x − + 4 4 8 ( 1) 4 x − + 8 3( 1)5 5 x − + 2 6 8 ( 1) 6 x − + 7 8 ( 1) 7 x − + 8 8 ( 1) 8 − = x8− 8x7+ 28x6− 56x5+ 70x4− 56x3+ 28x2 − 8x + 1 d (2x + 3)4= 4 (2 )4 0 x + 3 4 (2 ) 3 1 x + 2 2 4 (2 ) 3 2 x + 4 (2 )33 3 x + 4 4 3 4 = 16x4+ 96x3+ 216x2+ 216x + 81 e (7 − x)4 = 4 74 0 + 3 2 2 4 4 7 ( ) 7 ( ) 1 x 2 x − + − + 4 7( )3 4 ( )4 3 x 4 x − + − = 2401 − 1372x + 294x2− 28x3+ x4 f (2 − 3x)5 = 5 25 0 + 4 5 2 ( 3 ) 1 x − + 3 2 5 2 ( 3 ) 2 x − + 5 2 ( 3 )2 3 3 x − + 4 5 2( 3 ) 4 x − + 5 ( 3 )5 5 x − = 32 − 240x + 720x2 − 1080x3 + 810x4 − 243x5 2 a 3 1 x x + = x 3+ 3x21 x + 3x 2 1 x + 3 1 x = x3+ 3x + 3 x + 3 1 x b 7 2 3x x − = (3x) 7− 7(3x)62 x + 21(3x)5 2 2 x − 35(3x)4 3 2 x + 35(3x)3 4 2 x − 21(3x)2 5 2 x + 7(3x) 6 7 2 2 x x − = 2187x7 − 10 206x5 + 20 412x3 − 22 680x + 3 15120 6048 x − x + 5 7 1344 128 x − x c 6 2 3 x x + = (x 2)6+ 6(x2)53 x + 15(x2)4 2 3 x + 20(x2)3 3 3 x + 15(x2)2 4 3 x + 6(x2) 3 5 x + 6 3 x = x12+ 18x9+ 135x6+ 540x3+ 1215 + 3 1458 x + 6 729 x d 5 2 3 2x x − = 5 4 2 2 3 5 3 2x x x − + 3 2 2 3 10 (2 )x x − 2 3 2 3 10 (2 )x x + 4 5 2 3 5 (2 )x (2 )x x − = 10 7 243 810 x − x + 4 1080 720 x x − + 240x2− 32x5 3 (r + 1)th term is n r (ax) n − rbr a (x − 7)3 i x2 is the 2nd term ⇒ r = 1 Coefficient = 13 x 2(−7)1 = −21 ii x3 is first term ⇒ r = 0 term = 3 0 x 370 Coefficient = 1 iii x4 Coefficient = 0 b (2x + 1)5 i x2 is the 4th term ⇒ r = 3 term = 53 (2x) 213 Coefficient = 40 ii x3 is the third term
⇒ r = 2 term = 5 2 (2x) 312 Coefficient = 80 iii x4 is the 2nd term ⇒ r = 1
term = 15 (2x) 411 Coefficient = 80 c 5 2 3x x + i Coefficient of x2= 0
M M 1 2 - 1
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G r a p h s a n d p o l y n o m i a l s ii x3 is the 5th term ⇒ r = 4 term = 1 5 2 4 x (3x) 4 Coefficient = 810 iii Coefficient of x4= 0 d 6 2 3 x x − i Coefficient of x2= 0 ii x3 is the 4th term ⇒ r = 3 term = 63 (x 2)3 3 3 x − Coefficient = −540 iii Coefficient of x4= 0 e 6 2 3 7x x + i Coefficient of x2= 0 ii x3 is the 2nd term ⇒ r = 1 term = 16 (7x) 5 1 2 3 x Coefficient = 302 526 iii Coefficient of x4 is 0. 4 3 2 5 3x x − x3 is the 2nd term ⇒ r = 1 term = 13 (3x 2)2 1 5 x − = 3 × 9x4× 1 5 x − = −135x3 The answer is A.5 When the expression for C is expanded it does not contain an x5 term. The first three terms contain x8, x6 and x4
respectively. All the other expressions contain an x5 term.
The answer is C. 6 5 3 2 2 x x + = (x 3)5 + 5(x3)4 2 2 x + 10(x3)3 2 2 2 x + 10(x3)2 3 2 2 x + 5(x3) 4 2 2 x + 5 2 2 x = 1x15+ 10x10+ 40x5+ 80 + 3 80 x + 2 32 x ∴ = 1 + 10 + 40 + 80 + 80 + 32 = 243 The answer is D. 7 (2x − 3)4 = (2x)4− 4(2x)33 + 6(2x)232− 4(2x)33+ 34 = 16x4− 96x3+ 216x2− 216x3+ 81 The answer is D. 8 Fourth term = 6C 3x3× (3y)3 = 20 × x3 × 27y3 = 540x3y3 9 Term 3 ⇒ r = 2 = 2 7 9 3 2 4 x − = 78 73216 x2 = 19 683 2 4 x 10 x6, x3, x0 3rd term is independent of x. r = 2 = 2 4 2 6 (3 ) 2 2 x x = 4860 11 Powers of x are (x2)5, (x2)4 13 x , (x2)3 2 3 1 x x10, x5, x0,
The third term is independent of x. term = 5C 2(x2)3 2 3 4 x − = 10 × +16 = 160 12 4 2 2 3 x x + Powers of x are (x2)4, (x2)3 12 x , (x2)2 2 2 1 x x8, x4, x0
The third term is independent of x. term = 4C 2(x2)2 2 2 3 x = 6 × 9 = 54 13 Expand (p + 3)5 = p5+ 5p43 + 10p332+ 10p233+ 5p34+ 35 = p5+ 15p4+ 90p3+ 270p2+ 405p + 243 ∴(2p − 5)(p5+ 15p4+ 90p3+ 270p2+ 405p + 243) ⇒ 2p6+ 30p5+ 180p4+ 540p3+ 810p2+ 486p − 5p5− 75p4− 450p3− 1350p2− 2025p − 1215 Coefficient of 4th term = 180 − 75 = 105 14 (2a − 1)n 2nd term is nC 1(2a)n − 1(−1)1 coefficient: −n × 2n − 1= −192 n × 2n × 12 = 192 n × 2n= 384 = 3 × 27 = 3 × 2 × 26 = 6 × 26 n = 6
Exercise 1B — Polynomials
1 Polynomial expressions consist of terms which have non– negative integer powers of x only.
Not Polynomial: ii x4+ 3x2− 2x + x iii x7+ 3x6− 2xy + 5x vi 2x5+ x4− x3+ x2+ 3x − 2 x Polynomial: i x3− 2x iv 3x8− 2x5+ x2− 7 v 4x6− x3+ 2x − 3 2 a P(x) + Q(x) = 8 − 3x + 2x2+ x4+ x5− 3x4− 4x2− 1 = x5− 2x4− 2x2− 3x + 7 b Q(x) − R(x) = x5− 3x4− 4x2− 1 − (8x3+ 7x2− 4x) = x5− 3x4− 4x2− 1 − 8x3− 7x2+ 4x = x5− 3x4− 8x3− 11x2+ 4x − 1 c 3P(x) − 2R(x). 3P(x) = 3(8 − 3x + 2x2 + x4) = 24 − 9x + 6x2 + 3x4
G r a p h s a n d p o l y n o m i a l s M M 1 2 - 1
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2R(x) = 2(8x3+ 7x2− 4x) = 16x3+ 14x2− 8x ∴ 3P(x) − 2R(x) = 24 − 9x + 6x2+ 3x4 − (16x3+ 14x2− 8x) = 24 − 9x + 6x2+ 3x4− 16x3− 14x2+ 8x = 3x4− 16x3− 8x2− x + 24 d 2P(x) − Q(x) + 3R(x) 2P(x) = 2(8 − 3x + 2x2+ x4) = 16 − 6x + 4x2+ 2x4 3R(x) = 3(8x3 + 7x2 − 4x) = 24x3+ 21x2− 12x 2P(x) − Q(x) + 3R(x) = 16 − 6x + 4x2+ 2x4− (x5− 3x4− 4x2− 1) + 24x3+ 21x2− 12x = 16 − 6x + 4x2+ 2x4− x5+ 3x4+ 4x2+ 1 + 24x3+ 21x2− 12x = 17 − 18x + 29x2+ 24x3+ 5x4− x5 3 a P(x) = x6+ 2x5− x3+ x2 i degree = 6 ii P(0) = 06+ 2 × 05− 03+ 02 = 0 iii P(2) = 26+ 2 × 25− 23+ 22 = 124 iv P(−1) = −16+ 2 × −15− (−1)3+ (−1)2 = 1 b P(x) = 3x7− 2x6+ x5− 8 i degree = 7 ii P(0) = 3 × 07− 2 × 06+ 05− 8 = −8 iii P(2) = 3 × 27− 2 × 26+ 25− 8 = 280 iv P(−1) = 3 × (−1)7− 2 × (−1)6+ (−1)5− 8 = −3 − 2 − 1 − 8 = −14 c P(x) = 5x6+ 3x4− 2x3− 6x2+ 3 i degree = 6 ii P(0) = 5 × 06+ 3 × 04− 2 × 03− 6 × 02+ 3 = 3 iii P(2) = 5 × 26+ 3 × 24− 2 × 23− 6 × 22+ 3 = 331 iv P(−1) = 5 × (−1)6+ 3 × (−1)4 − 2 × (−1)3− 6 × (−1)2+ 3 = 5 + 3 + 2 − 6 + 3 = 7 d P(x) = −7 + 2x − 5x2+ 2x3− 3x4 i degree = 4 ii P(0) = −7 + 2 × 0 − 5 × 02+ 2 × 03− 3 × 04 = −7 iii P(2) = −7 + 2 × 2 − 5 × 22 + 2 × 23 − 3 × 24 = −55 iv P(−1) = −7 + 2 × (−1) − 5 × (−1)2 + 2 × (−1)3− 3(−1)4 = − 7 − 2 − 5 − 2 − 3 = −19 4 P(x) = x8− 3x6+ 2x4− x2+ 3 P(−2) = (−2)8− 3 × (−2)6+ 2 × (−2)4− (−2)2+ 3 = 95 The answer is B. 5 P(x) = 2x7+ ax5+ 3x3+ bx − 5 P(1) = 4 ∴ 4 = 2 × 17+ a × 15+ 3 × 13+ b × 1 − 5 4 = 2 + a + 3 + b − 5 4 = a + b [1] P(2) = 163 163 = 2 × 27+ a × 25+ 3 × 23+ b × 2 − 5 = 256 + 32a + 24 + 2b − 5 −112 = 32a + 2b [2] [1] × 2 8 = 2a + 2b [3] [2] − [3] −120 = 30a a = −4 b = 8 6 f(x) = ax4+ bx3− 3x2− 4x + 7 f(1) = −2 ∴ −2 = a × (1)4+ b × (1)3− 3 × 12−4 × 1 + 7 −2 = a + b − 3 − 4 + 7 −2 = a + b ∴ −2 − b = a [1] f(2) = −5 ∴ −5 = a × 24+ b × 23− 3 × 22− 4 × 2 + 7 −5 = 16a + 8b − 12 − 8 + 7 −5 = 16a + 8b − 13 8 = 16a + 8b 8 = 8(2a + b) 1 = 2a + b [2] Substitute [1] into [2] 1 = 2(−2 − b) + b 1 = −4 − 2b + b 1 = −4 − b b = −5 If b = −5, then [1] −2 − −5 = a. 3 = a ∴ f(x) = 3x4 − 5x3 − 3x2 − 4x + 7 7 Q(x) = x5 + 2x4 + ax3 − 6x + b Q(2) = 45 ∴ 45 = 25 + 2 × 24 + 23a − 6 × 2 + b 45 = 52 + 8a + b −7 = 8a + b −7 − 8a = b [1] Q(0) = −7 ∴ −7 = 05 + 2 × 04 + a × 03 − 6 × 0 + b −7 = b [2] Substitute [2] into [1]. −7 − 8a = −7 −8a = 0 a = 0 ∴ Q(x) = x5 + 2x4 − 6x − 7. 8 P(x) = ax6 + bx4 + x3 − 6 If 3P(1) = −24 then 3P(x) = 3(ax6 + bx4 + x3 − 6) −24 = 3(a × 16 + b × 14 + 13 − 6) −8 = a + b + 1 − 6 −8 = a + b − 5 −3 = a + b −3 − a = b [1] If 3P(−2) = 102 then 3P(x) = 3(ax6 + bx4 + x3 − 6) 102 = 3[a(−2)6 + b(−2)4 + (−2)3 − 6) 34 = 64a + 16b − 8 − 6 34 = 64a + 16b − 14 48 = 64a + 16b (÷ 16) 3 = 4a + b [2] Substitute [1] into [2] 3 = 4a + (−3 − a) 3 = 4a − 3 − a 6 = 3a 2 = a If 2 = a then b = −3 − a b = −3 − 2 b = −5 ∴ P(x) = 2x6 − 5x4 + x3 − 6 9 a P(x) = ax4 − x3 + 3x2 − 5 If P(1) = −1 then −1 = a × (1)4 − (1)3 + 3 × (1)2 − 5M M 1 2 - 1
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G r a p h s a n d p o l y n o m i a l s −1 = a − 1 + 3 − 5 −1 = a − 3 2 = a The answer is C. b f(x) = xn − 2x3 + x2 − 5x If f(2) = 10 then 10 = 2n − 2 × 23 + 22 − 5 × 2 10 = 2n − 16 + 4 − 10 10 = 2n − 22 32 = 2n 25 = 2n ∴ n = 5 The answer is D.Exercise 1C — Division of polynomials
1 a 2 3 2 3 2 2 2 2 13 2 5 2 4 4 2 5 2 8 13 2 13 52 50 x x x x x x x x x x x x x x + + − + − − − − + − − − − − Q(x) = x2 + 2x + 13 R(x) = 50 b 5 4 3 2 5 4 4 3 4 3 3 2 3 2 2 2 0 3 0 4 3 3 3 3 3 3 9 6 0 6 18 18 4 18 54 58 3 58 174 x x x x x x x x x x x x x x x x x x x x x x + − + + + + − + − − − − − + − + − + − − − + − + 4 3 3 6 2 18 58 171 x − x + x − x+ − Q(x) = x4 − 3x3 + 6x2− 18x + 58 R(x) = −171 c 3 4 3 2 4 3 3 2 3 2 2 6 17 6 2 4 0 3 6 18 17 2 17 51 53 4 53 159 155 0 155 465 465 x x x x x x x x x x x x x x x x x x x + − + − + − − − + − − − − − + − − 2+53x+155 Q(x) = 6x3 + 17x2 + 53x + 155 R(x) = 465 d 3 2 4 3 2 4 3 3 2 3 2 7 7 101 3 9 27 3 6 0 12 0 3 1 3 7 0 7 7 3 x x x x x x x x x x x x x x − + + − + + + + − + − + − − − 2 2 7 12 3 7 7 3 9 101 0 9 101 101 9 27 101 27 x x x x x x + − + + − + − Q(x) = x3 − 7 3x 2 + 7 9x + 101 27 R(x) = 320 27 − =−27101 2 a i P(x) = x3 − 2x2 + 5x − 2 P(4) = 43 − 2 × 42 + 5 × 4 − 2 = 50 ii P(x) = x4 + x3 + 3x2 − 7x P(1) = 1 + 1 + 3 − 7 = −2 iii P(x) = x5 − 3x3 + 4x + 3 P(−3) = (−3)5 − 3 × (−3)3 + 4 × (−3) + 3 = −171 iv P(x) = 2x6 − x4 + x3 + 6x2 − 5x P(−2) = 2 × (−2)6 − (−2)4 + (− 2)3+ 6 × (−2)2 − 5 × (−2) = 138 v P(x) = 6x4 − x3 + 2x2 − 4x P(3) = 6 × 34 − 33 + 2 × 32 − 4 × 3 = 465 vi P(x) = x4 − 13x2 + 36 P(2) = 24 − 13 × 22 + 36 = 0 vii P(x) = 3x4 − 6x3 + 12x 1 3 P− = 3 × 4 1 3 − − 6 × 3 1 3 − + 12 × − 1 3 = −320 27 viii P(x) = x5 + 3x3 − 4x2 + 6x − 8 3 2 P = 5 3 2 + 3 × 3 3 2 − 4 × 2 3 2 + 6 × 3 2 − 8 = 923 32
b The values obtained in 2 were the same as the remainder values obtained in 1.
3 a P(3) = 33 + 9 × 32 + 26 × 3 − 30
= 156
Since P(3) ≠ 0, x − 3 is not a factor
b P(−2) = (−2)4 − (−2)3 − 5 × (−2)2 − 2 × (−2) − 8
= 0
G r a p h s a n d p o l y n o m i a l s M M 1 2 - 1
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c P(+ 4) = 4 − 9 × + 4 + 6 × 42 − 13 × (+ 4)3 − 12 × (+ 4)4 + 3 × (+ 4)5 = 4 − 36 + 94 − 832 − 3072 + 3072 = −768Since P(−4) ≠ 0 then 4 − x is not a factor. d P 1 2 − = 4 × 6 1 2 − + 2 × 5 1 2 − − 8 × 4 1 2 − − 4 × 3 1 2 − + 6 × 2 1 2 − − 9 × − 1 2 − 6 = 0.0625 + − 0.0625 − 0.5 + 0.5 + 1.5 + 4.5 − 6 = 0 Since P 1 2 − = 0 then 2x + 1 is a factor. 4 a f(x) = x4 − 4x3 − x2 + 16x − 12 A x + 1 ⇒ f(− 1) = (−1)4 − 4 × (−1)3 − (−1)2 + 16 × (−1) − 12 = 1 + 4 − 1 − 16 − 12 = −24 B x ⇒ f(0) = −12 C x + 2 ⇒ f(− 2) = (−2)4 − 4 × (−2)3 − (−2)2 + 16 × (−2) − 12 = 16 + 32 − 4 − 32 − 12 = 0
Since f(−2) = 0 then (x + 2) is a factor.
D x + 3 ⇒ f(−3) = (−3)4 − 4 × (−3)3 − (−3)2 + 16 × (−3) − 12 = 120 E x − 4 ⇒ f(4) = 44 − 4 × 43 − 42 + 16 × 4 − 12 = 36 The answer is C. b 3 2 4 3 2 4 3 3 2 3 2 2 2 6 11 6 4 16 12 2 2 6 6 12 11 16 11 22 6 12 6 12 0 x x x x x x x x x x x x x x x x x x x x − + − − − + − + − + − − − − − + − + − − − − − Test x = 1 into x3 − 6x2 + 11x − 6 = 13 − 6 × 1 + 11 − 6 = 0 ∴ x − 1 is a factor. 2 3 2 3 2 2 2 5 6 6 11 6 1 5 11 5 5 6 6 6 6 0 x x x x x x x x x x x x x x − + − + − − − − − + − − + − − − x2− 5x + 6 = (x − 3)(x − 2) f(x) factorises to (x + 2)(x − 1)(x − 3)(x − 2) The answer is B. 5 a P(x) = x3 + 4x2 − 3x − 18 Test x = ± 1 P(x) ≠ 0 x = 2, P(x) = 0 ∴ (x − 2) is a factor 2 3 2 3 2 2 2 6 9 4 3 18 2 2 6 3 6 12 9 18 9 18 0 x x x x x x x x x x x x x x + + + − − − − − − − − − − − ∴ (x − 2)(x2 + 6x + 9) = (x − 2)(x + 3)2 b P(x) = 3x3 − 13x2− 32x + 12 Test x = ± 1 P(x) ≠ 0 x = ± 2 when x = −2, P(x) = 0 ∴ (x + 2) is a factor 2 3 2 3 2 2 2 3 19 6 3 13 32 12 2 3 6 19 32 19 38 6 12 6 12 0 x x x x x x x x x x x x x x + − + − − + + − + − − − − − + − + ∴ (x + 2)(3x2 − 19x + 6) = (x + 2)(3x − 1)(x − 6) c P(x) = x4 + 2x3 − 7x2 − 8x + 12 Test x = −2, P(x) = 0 ∴ (x + 2) is a factor 3 2 4 3 2 4 3 2 2 0 7 6 2 7 8 12 2 2 0 7 8 7 14 6 12 6 12 0 x x x x x x x x x x x x x x x x + − + + − − + + − + − − − − − + − + ∴ (x + 2)(x3 + 0x2 − 7x + 6) Test x = 2, P(x) = 0 ∴ (x − 2) is a factor 2 3 2 3 2 2 2 2 3 0 7 6 2 2 2 7 2 4 3 6 3 6 0 x x x x x x x x x x x x x x + − + − + − − − − − − − + −− + ∴ (x + 2)(x − 2)(x2 + 2x − 3) (x + 2)(x − 2)(x + 3)(x − 1)
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G r a p h s a n d p o l y n o m i a l s d P(x) = 4x4 + 12x3 − 24x2 − 32x Test x = −1, P(x) = 0 ∴ x + 1 is a factor 3 2 4 3 2 4 3 3 2 3 2 2 2 4 8 32 4 12 24 32 0 1 4 4 8 24 8 8 32 32 32 32 0 x x x x x x x x x x x x x x x x x x + − + − − + + − + − − + − − − − − ∴ (x + 1)(4x3 + 8x2 − 32x)Take out factor of 4x. 4x(x + 1)(x2+ 2x − 8) ∴ 4x(x + 1)(x − 2)(x + 4) 6 a 3x3 + 3x2 − 18x = 0 Test x = 2, f(x) = 0 ∴ (x − 2) is a factor 2 3 2 3 2 2 2 3 9 3 3 18 0 2 3 6 9 18 9 18 0 x x x x x x x x x x x x + + − + − − − − − − ∴ (x − 2)(3x2 + 9x) = 0 3x(x − 2)(x + 3) = 0 ∴ x = 0, 2, or −3 b 2x4 + 10x3 − 4x2 − 48x = 0 Test x = 2, f(x) = 0 ∴ (x − 2) is a factor 3 2 4 3 2 4 3 3 2 3 2 2 2 2 14 24 2 10 4 48 2 2 4 14 4 14 28 24 48 24 48 0 x x x x x x x x x x x x x x x x x x + + + − − − − − − − − − − − ∴ (x − 2)(2x3 + 14x2 + 24x)
Take out common factor of 2x: 2x(x − 2)(x2+ 7x + 12) = 2x(x − 2)(x + 3)(x + 4) = 0 ∴ x = 2, −3, 0, and −4 c 2x4 + x3 − 14x2 − 4x + 24 = 0 Test x = 2, f(x) = 0 ∴ (x − 2) is a factor 3 2 4 3 2 4 3 3 2 3 2 2 2 2 5 4 12 2 14 4 24 2 2 4 5 14 5 10 4 4 4 8 x x x x x x x x x x x x x x x x x x + − − + − − + − − − − − − − − − − + 12 24 12 24 0 x x − + − − + ∴ (x − 2)(2x3 + 5x2 − 4x − 12) Test x = −2, f(−2) = 0 ∴ (x + 2) is a factor 2 3 2 3 2 2 2 2 6 2 5 4 12 2 2 4 4 2 6 12 6 12 0 x x x x x x x x x x x x x x + − + − − + − + − − − − − − − − ∴ (x − 2)(x + 2)(2x2 + x − 6) (x − 2)(x + 2)(2x − 3)(x + 2) x = 2, −2, or 32 d x4 − 2x2 + 1 = 0 Test x = + 1, f(x) = 0 ∴ (x − 1) is a factor 3 2 4 3 2 4 3 3 2 3 2 2 2 1 0 2 0 1 1 2 0 1 1 0 x x x x x x x x x x x x x x x x x x x x + − − + − + + − − − − − − − + − − + − + − − + ∴ (x − 1)(x3 + x2 − x − 1) Test x = −1, f(x) = 0 ∴ (x + 1) is a factor 2 3 2 3 2 1 1 1 0 1 1 0 x x x x x x x x x − + − − + − + − − − − − ∴ (x − 1)(x + 1)(x2 − 1) = 0 (x − 1)(x + 1)(x − 1)(x + 1) = 0 x = ±1 7 If (x − 2) is a factor then when x = 2, f(x) = 0 0 = x3 + ax2 − 6x − 4 f(2) = 0 = 23 + a22 − 6 × 2 − 4 0 = 8 + 4a − 12 − 4 0 = 4a − 8 8 = 4a 2 = a 8 Let P(x) = x3 + x2 − ax + 3 P(1) = 1 + 1 − a + 3 = 0 (x − 1) is a factor a = 5
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9 If (x + 3) is a factor then when x = −3, f(x) = 0 f(−3) = 0 = 2(−3)4 + a(−3)3 − 3 × (−3) + 18 0 = 162 − 27a + 9 + 18 0 = 189 − 27a 27a = 189 a = 7 10 If (x + 1) is a factor then when x = −1, f(x) = 0 f(−1) = 0 = −a − 4 − b − 12 0 = −a − b − 16 a = −b − 16 [1] If (x − 2) is a factor then when x = 2, f(x) = 0 f(2) = 0 = 8a − 16 + 2b − 12 0 = 8a + 2b − 28 28 = 8a + 2b 14 = 4a + b [2] Sub [1] into [2] 14 = 4(−b − 16) + b 14 = −4b − 64 + b 14 = −3b − 64 78 = −3b −26 = b ∴ a = + 26 − 16 a = 1011 (2x − 3) and (x + 2) are factors of 2x3 + ax2 + bx + 30 P(x) = 2x3 + ax2 + bx + 30 P(−2) = 2(−2)3 + a(−2)2 + b(−2) + 30 = 0 −16 + 4a − 2b + 30 = 0 4a − 2b = −14 [1] P 3 2 = 2 3 3 2 + a 2 3 2 + b 3 2 + 30 = 0 2 × 27 84 + 9 4 a + 3 2 b + 30 = 0 27 + 9a + 6b + 120 = 0 9a + 6b = −147 3a + 2b = −49 [2] [1] + [2] 7a = −63 a = −9 Substitute into [1] 4 × −9 − 2b = −14 −2b = 22 b = −11 a = −9, b = −11.
Exercise 1D — Linear graphs
1 a 2x + 3y = 12 x-intercept when y = 0 2x = 12 x = 6 y-intercept when x = 0 3y = 12 y = 4 b 2y − 5x − 10 = 0 x-intercept when y = 0 −5x = 10 x = −2 y-intercept when x = 0 2y = 10 y = 5 c 2x − y = 1 x-intercept when y = 0 2x = 1 x = 1 2 y-intercept when x = 0 −y = 1 y = −1 2 a y = mx + c y = 3x + c find c in (2, 1) 1 = 3 × 2 + c −5 = c ∴ y = 3x − 5 −3x + y + 5 = 0 b y = mx + c y = −2x + c find c, sub in (−4, 3) 3 = −2 × −4 + c 3 = 8 + c −5 = c ∴ y = −2x − 5 2x + y + 5 = 0 3 a (−3, −4), (−1, −10) m = 10 4 1 3 − + − + = 6 2 − = −3 y = −3x + c sub in (−3, −4) −4 = −3 × −3 + c −13 = c y = −3x − 13 3x + y + 13 = 0 b (7, 5), (2, 0) m = 5 0 7 2 − − = 5 5 = 1 y = x + c sub in (2, 0) 0 = 2 + c −2 = c y = x − 2 −x + y + 2 = 0 4 2y − 3x − 6 = 0 A 2 × 6 − 3 × 2 − 6 = 0 12 − 6 − 6 = 0 B 2 × 0 − 3 × − 2 − 6 = 0 0 + 6 − 6 = 0 C 2 × 3 − 3 × 0 − 6 = 0 6 − 0 − 6 = 0 D 2 × 2 − 3 × 1 − 6 = 0 4 − 3 − 6 ≠ 0 E 2 × 9 − 3 × 4 − 6 = 0 18 − 12 − 6 = 0 The answer is D. 5 a i −2 = 5 1 2 b− + −2 = b3−5 −6 = b − 5 −1 = b ii y − x = 7 ∴ y = x + 7 m = 1 1 = 5 1 2 b− + 3 = b − 5 8 = b b parallel to y = 3x − 4 ∴ m = 3 y = 3x + c sub in (4, 5) 5 = 3 × 4 + c −7 = c ∴ y = 3x − 7 0 = 3x − y − 7 c 2y − x + 1 = 0 2y = x − 1 y = 1 2x − 1 2
m = −2 gradient of perpendicular line y − y1= m(x − x1) Sub in (−2, 4) y − 4 = −2(x + 2) y − 4 = −2x − 4 2x + y = 0 6 i x + 2y + 4 = 0 • x-intercept when y = 0 x = − 4 • y-intercept when x = 0 2y = − 4 y = − 2 Graph e ii y = 3 − Graph f iii y − 2x − 2 = 0 • x-intercept when y = 0 −2 = 2x −1 = x • y-intercept when x = 0 y = 2 Graph a iv 3y + 2x = 6 • x-intercept when y = 0 2x = 6 x = 3 • y-intercept when x = 0 3y = 6 y = 2 Graph c v y − 2x = 0
• x- and y-intercepts occur at the origin.
Graph b vi x = −2 − Graph d.
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G r a p h s a n d p o l y n o m i a l s 7 a y ≥ −2 or [−2, ∞) b y ≥ −5 or (−5, ∞) c −2 ≤ y < 3 or [−2, 3) d −2 ≤ y ≤ 3 or [−2, 3] e R f −∞ < y < 6 or (−∞, 6) 8 a 4y + 3x = 24 x ∈ [−12, 12] x-intercept 3x = 24 x = 8 y-intercept 4y = 24 y = 6 when x = −12 y = 15 when x = 12 y = −3 i domain [−12, 12] ii range [−3, 15] b 2x − 5y = 10x < 5 x-intercept 2x = 10 x = 5 y-intercept −5y = 10 y = −2 when x = 5, y = 0 i domain (−∞, 5) ii range (−∞, 0) c 4x − 3y − 6 = 0 x ∈ [2, 5) x-intercept 4x = 6 x = 3 2 y-intercept −3y = 6 y = −2 when x = 2 y = 2 3 when x = 5 y = 143 i Domain [2, 5) ii Range 2 14, 3 3 9 a Parallel ∴ m = −2 y = −2x + c sub in (2, 5) 5 = −4 + c 9 = c ∴ y = −2x + 9 2x + y − 9 = 0 b Perpendicular ∴ m = −13 y = −13x + c sub in (2, 5) (y − 5) = −1 3 (x − 2) y − 5 = 3 x − + 2 3 3y − 15 = −x + 2 3y + x − 17 = 0 x + 3y − 17 = 0 10 a Parallel to 4x − 13 = 2y 2x − 13 2 = y m = 2 ∴ y = 2x + c sub in (−3, 1) 1 = 2 × −3 + c 7 = c y = 2x + 7 −2x + y − 7 = 0 b 4x − 2y = 13 4x − 13 = 2y 2x − 13 2 = y ∴ m = 2 Perpendicular m = −1 2 sub in (−3, 1) y − 1 = −1 2(x + 3) y − 1 = −12x − 32 y = −12x − 12 2y = −x − 1 x + 2y + 1 = 0 11 3x − y = −2 3x + 2 = y m = 3 ax + 2y = 3 2y = −ax + 3 y = 2 a − x + 3 2 ∴ 3 = 2 a − 6 = −a −6 = a The answer is E. 12 5x + y − 3 = 0 bx − y − 2 = 0 y = −5x + 3 y = bx − 2 gradient −5 gradient b = 1 5 The answer is B.Exercise 1E — Quadratic
graphs
1 b2 − 4ac = ∆ a f(x) = x2− 3x + 4 a = 1, b = −3, c = 4 ∆ = 9 − 16 ∆ = −7 b2− 4ac < 0. No x-intercepts b f(x) = x2+ 5x − 8 a = 1 b = 5 c = −8 ∆ = 25 + 32 = 57b2− 4ac > 0. Two x-intercepts
c f(x) = 3x2− 5x + 9 a = 3 b = −5 c = 9 ∆ = 25 − 108 ∆ = −83 b2− 4ac < 0. No x-intercepts d f(x) = 2x2+ 7x − 11 a = 2 b = 7 c = −11 ∆ = 49 + 88 = 137
b2− 4ac > 0. Two x-intercepts e f(x) = 1 − 6x − x2
a = −1 b = −6 c = 1 ∆ = 36 + 4 = 40
b2
− 4ac > 0. Two x-intercepts f f(x) = 3 + 6x + 3x2
a = 3 b = 6 c = 3
∆ = 36 − 36 = 0
b2
− 4ac = 0. One x-intercept 2 a f(x) = x2 − 6x + 8 y-intercept x = 0 y = 8 x-intercept(s) 0 = (x − 4)(x − 2) ∴ x = 4 or 2 b f(x) = x2− 5x + 4 y-intercept x = 0 y = 4 x-intercept y = 0 0 = (x − 4)(x − 1) x = 4 or 1 c f(x) = 10 + 3x − x2 y-intercept x = 0 y = 10 x-intercept y = 0 0 = (5 − x)(2 + x) x = 5 or −2 d f(x) = 6x2− x − 12 y-intercept x = 0 y = −12
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x-intercept(s) y = 0 0 = (3x + 4) (2x − 3) x = −4 3 or 3 2 3 a f(x) = x2− 6x + 8 = x2− 6x + 32− 32+ 8 = (x − 3)2− 9 + 8 = (x − 3)2− 1 TP is (3, −1) b f(x) = x2− 5x + 4 = x2− 5x + 5 2 5 2 2 2 − + 4 = 2 5 2 x − − 25 4 + 4 = 2 5 2 x − − 25 4 + 16 4 = 2 5 2 x − − 9 4 TP is 5, 9 2 4 − c f(x) = 10 + 3x − x2 = −(x2− 3x − 10) = − 2 2 2 3 3 3 10 2 2 x x − + − − = − 2 3 9 10 2 4 x − − − = − 2 3 9 40 2 4 4 x − − − = − 2 3 2 x − + 49 4 TP is 3, 49 2 4 + = 1 1 1 , 12 2 4 d f(x) = 6x2− x − 12 = 6 2 2 6 x x − − = 2 2 2 1 1 6 2 6 12 12 x x − + − − = 2 1 1 288 6 12 144 144 x − − − = 2 1 289 6 12 144 x − − = 2 1 6 12 x − − 289 24 TP is 1, 12 1 12 24 4 a i y = 2 − x2 a = −1, h = 0, k = 2 TP = (0, 2) ii Domain = R iii Range = (−∞, 2] b i y = (x − 6)2 a = 1, h = 6, k = 0 TP = (6, 0) ii Domain = R iii Range = [0, ∞) c i y = −(x + 2)2 a = −1, h = −2, k = 0 TP = (−2, 0) ii Domain = R iii Range = (−∞, 0] d i y = 2(x + 3)2− 6 a = 2, h = −3, k = −6 TP = (−3, −6) ii Domain = R iii Range = [−6, ∞) 5 Using y = A (x + B)2 + C a i TP = (1, − 2) ∴ B = −1 and C = −2 Assume A = 1 ⇒ y = 1(x − 1)2 − 2 y = x2− 2x + 1 − 2 y = x2− 2x − 1 ii Domain = R iii Range [−2, ∞) b i TP = (2, −3) ∴ B = −2 C = −3 Assume A = 1 ⇒ y = 1 (x − 2)2 − 3 = x2− 4x + 4 − 3 = x2 − 4x + 1 ii Domain = [−1, ∞) iii Range = [−3, ∞) c i TP = (1, 9) ∴ B = −1 and C = 9 Assume A = −1 ⇒ y = −1(x − 1)2+ 9 y = −1(x2 − 2x + 1) + 9 y = −x2+ 2x − 1 + 9 y = −x2 + 2x + 8 ii Domain = [−4, 4) iii Range = [−16, 9] 6 a y = 2x2+ 3 TP = (0, 3) y-intercept when x = 0 y = 3 x-intercepts when y = 0 0 = 2x2+ 3There are no x-intercepts.
b y = 1 − 4(2 − x)2 TP = (2, 1) y-intercept when x = 0 y = 1 − 4 × 4 = −15 x-intercepts when y = 0 0 = 1 − 4(2 − x)2 0 = 1 − 4(4 − 4x + x2) = 1 − 16 + 16x − 4x2 = −4x2+ 16x − 15 x = 3 2 and x = 5 2 c y = (2x − 3)2 − 8 TP = 3, 8 2 − y-intercept when x = 0 y = (−3)2− 8 = 1 x-intercepts when y = 0 0 = (2x − 3)2− 8 = 4x2− 12x + 9 − 8 = 4x2− 12x + 1
From the graphics calculator,
x = 2.91 and x = 0.09 7 a y = x2− 2x − 3 x-intercepts y = 0 0 = (x − 3)(x + 1) x = 3 or −1 (3, 0)(−1, 0) The answer is B. b y = x2 − 2x − 3 y = x2− 2x + 12− 12− 3 y = (x − 1)2 − 4 TP = (1, −4) The answer is C. 8 f(x) = −(x + 3)2+ 4 TP = (−3, 4) ∴ range (−2, 4] The answer is D. 9 y = (x − 4)2 x ∈ [0, 6] TP = (4, 0) When x = 0 y = (−4)2 = 16 ∴ range [0, 16] When x = 6 y = (6 − 4)2 = 22 = 4 But x = 4 y = 0 The answer is A. 10 a f(x) = (x − 2)2− 4 TP = (2, −4) y-int x = 0 y = (0 − 2)2− 4 y = (−2)2− 4 y = 0
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G r a p h s a n d p o l y n o m i a l s x-int y = 0 0 = (x − 2 − 2)(x − 2 + 2) 0 = (x − 4)(x) ∴ x = 4 or 0 b f(x) = −(x + 4)2 + 9 TP = (−4, 9) y int x = 0 y = −(0 + 4)2+ 9 y = −16 + 9 y = −7 x-int y = 0 0 = 9 − (x + 4)2 0 = (3 − (x + 4))(3 + (x + 4)) 0 = (3 − x − 4) (3 + x + 4) 0 = (−x − 1)(7 + x) x = −1 or −7 c y = x2 + 4x + 3 y = x2+ 4x + 4 − 4 + 3 y = (x + 2)2− 1 TP = (−2, −1) y-int x = 0 y = 3 x-int y = 0 0 = (x + 2 − 1)(x + 2 + 1) 0 = (x + 1)(x + 3) ∴ x = −1 or −3 d y = 2x2− 4x − 6 y = 2[x2− 2x − 3] = 2[x2− 2x + 1 − 1 − 3] = 2[(x − 1)2− 4] y = 2(x − 1)2− 8 TP = (1, −8) y-int x = 0 y = −6 x-int y = 0 0 = 2[(x − 1 − 2)(x − 1 + 2)] 0 = 2(x − 3)(x + 1) ∴ x = 3 or −1 11 a y = x2− 2x + 2 x ∈ [−2, 2] y = x2− 2x + 1 − 1 + 2 y = (x − 1)2+ 1 ∴ TP = (1, 1) i Domain = [−2, 2] ii Range: When x = −2 y = 10 When x = 2 y = 2 but TP = (1, 1) ∴ [1, 10] b y = −x2+ x − 1 x ∈ R+ y = −(x2− x + 1) y = − 2 2 2 1 1 1 2 2 x x − + − + y = − 2 1 3 2 4 x − + y = − 2 1 2 x − − 3 4 ∴ TP = 1, 3 2 4 − i Domain = R+ ii Range = (−∞, −34] c f(x) = x2− 3x − 2 x ∈ [−10, 6] f(x) = x − 3x + 2 3 2 − 2 3 2 − 2 = 2 3 2 x − − 9 4 − 8 4 = 2 3 2 x − − 17 4 ∴ TP = 3, 17 2 4 − i Domain = [−10, 6] ii Range: When x = −10 y = 128 ∴ 17 ,128 4 − d f(x) = −3x2+ 6x + 5 x ∈ [−5, 3) = 3 2 2 5 3 x x − − − = 3 2 2 1 1 5 3 x x − − + − − = 3 ( 1)2 8 3 x − − − = −3(x − 1)2+ 8 TP = (1, 8) i Domain = [−5, 3) ii Range: When x = −5, y = −100 ∴ [−100, 8] 12 V(t) = 2t2− 16t + 40 t ∈ [0, 10] V(t) = 2(t2 − 8t + 20) = 2[t2− 8t + 16 − 16 + 20] = 2[(t − 4)2 + 4] = 2(t − 4)2+ 8 ∴ TP = (4, 8) When t = 0 V(t) = 40 When t = 10 V(t) = 2 × 62+ 8 = 80 a minimum V = 8 m3 b maximum V = 80 m3 13 h(t) = −3t2+ 12t + 36 h(t) = −3[t2− 4t − 12] = −3[t2− 4t + 4 − 4 − 12] = −3[(t − 2)2− 16] = −3(t − 2)2+ 48 ∴ TP = (2, 48) a maximum height = 48 m b When h(t) = 0 0 = −3[(t − 2 − 4)(t − 2 + 4)] 0 = −3(t − 6)(t + 2) ∴ t = 6 or −2Since time ⇒ 6 seconds c Domain [0, 6] Range [0, 48]
14 a h(t) = t2− 12t + 48, t ∈ [0, 11]
The lowest point is the
y-coordinate of the turning point
h(t) = t2
− 12t + 36 − 36 + 48 = (t − 6)2+ 12
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TP = (6, 12)
Lowest point is 12 m above the ground.
b Time taken is the x-coordinate of the turning point.
t = 6 seconds
c Check the end points of the domain
h(0) = 48 h(11) = 112
− 12 × 11 + 48 = 37
The highest point above the ground is 48 m.
d Domain = [0, 11] Range = [12, 48] e
Exercise 1F — Cubic graphs
1 a Positive cubic so a = 1. Goes through origin so x is a factor. y = x(x − a)(x − b)
= x(x + 6)(x − 5) b Positive cubic in form y = a(x − m)(x − n)2
∴ a = 1, m = 1, n = −2 ∴ y = 1(x − 1)(x + 2)2
2 a Positive cubic in form y = (x − l)(x − m)(x − n)
l = −3, m = 1, n = 4 ∴ y = (x + 3)(x − 1)(x − 4) ∴ (v)
b Negative cubic in form y = a(x − m)(x − n)2
∴ a = −1, m = 5, n = − 2 ∴ y = −1(x − 5)(x + 2)2 y = (5 − x)(x + 2)2
(iv)
c Negative cubic in form y = a(x − l)(x − m)(x − n)
a = −1, l = −3, m = 1, n = 4 ∴ y = −1(x + 3)(x − 1)(x − 4)
y = (x + 3)(1 − x)(x − 4) (ii)
d Positive cubic in form y = a(x − t)3 a = 1, t = 3 ∴ y = (x − 3)3
(i)
e Positive cubic in form
y = a(x − l)(x − m)(x − n)
a = 1, l = −4, m = −2, n = 1 ∴ y = (x + 4)(x + 2)(x − 1) (vi)
f Positive cubic in form y = a(x − m)(x − n)2 a = 1, m = 5, n = −2 y = (x − 5)(x + 2)2
(viii)
g Negative cubic in form
y = a(x − t)3
a = −1, t = 3 ∴ y = −1(x − 3)3 y = (3 − x)3
(vii)
h Negative cubic in form y = a(x − l)(x − m)(x − n) a = −1, l = − 4, m = −2, n = 1 ∴ y = −1(x + 4)(x + 2)(x − 1) y = (x + 4)(x + 2)(1 − x) (iii) 3 a y = x3 + x2 − 4x − 4 y-intercept x = 0 y = −4
Factorise to find x-intercepts Test x = −1, y = 0 ∴ x + 1 is a factor 2 3 2 3 2 4 4 4 1 4 4 4 4 0 x x x x x x x x x − + − − + − + − − − − − ∴ y = (x + 1)(x2− 4) y = (x + 1)(x − 2)(x + 2) If y = 0, x = −1, 2, or −2 b y = 2x3 − 8x2 + 2x + 12 y-int x = 0 y = 12
Factorise to find x-intercepts Test x = −1 so y = 0 ∴ (x + 1) is a factor 2 3 2 3 2 2 2 2 10 12 2 8 2 12 1 2 2 10 2 10 10 12 12 12 12 0 x x x x x x x x x x x x x x − + − + + + − + − + − − + + − + ∴ y = (x + 1)(2x2 − 10x + 12) y = 2(x + 1)(x − 2)(x − 3) If y = 0, then x = −1, 2 or 3 c y = −2x3 + 26x + 24 y-int x = 0 y = 24
Factorise to find x-intercepts Test x = −1 so y = 0 ∴ (x + 1) is a factor. 2 3 2 3 2 2 2 2 2 24 2 0 26 24 1 2 2 2 26 2 2 24 24 24 24 0 x x x x x x x x x x x x x x − + + − + + + + − − − + − + + − + ∴ y = (x + 1)(−2x2 + 2x + 24) y = 2(x + 1)(−x − 3)(x − 4) If y = 0, then x = −1, −3 or 4. d y = −x3 + 8x2 − 21x + 18 y-int x = 0 y = 18 x-intercept Factorise: Test x = 3 so y = 0 ∴ (x − 3) is a factor. 2 3 2 3 2 2 2 5 6 8 21 18 3 3 5 21 5 15 6 18 6 18 0 x x x x x x x x x x x x x x − + − − + − + − − − + − − − − + − − + ∴ y = (x − 3)(−x2 + 5x − 6) = −(x − 3)(x2 − 5x + 6) y = −1(x − 3)(x − 3)(x − 2) ∴ x = 3 or 2 4 a x3 + 6x2 + 12x + 8 = y Test x = −2 so y = 0 ∴ (x + 2) is a factor 2 3 2 3 2 2 4 4 6 12 8 2 2 4 12 4 8 4 8 4 8 0 x x x x x x x x x x x x x x + + + + + + − + + − + + − + ∴ y = (x + 2)(x2 + 4x + 4) = (x + 2)(x + 2)2 y = (x + 2)3 The answer is B.
